Planar Support for Nonpiercing Regions and Applications
E S A
Planar Support for Nonpiercing Regions and Applications
Rajiv Raman IIITDelhi 0
Delhi 0
India 0
0 Saurabh Ray Department of Computer Science , NYU Abu Dhabi , United Arab Emirates
Given a hypergraph H = (X, S), a planar support for H is a planar graph G with vertex set X, such that for each hyperedge S ∈ S, the subgraph of G induced by the vertices in S is connected. Planar supports for hypergraphs have found several algorithmic applications, including several packing and covering problems, hypergraph coloring, and in hypergraph visualization. The main result proved in this paper is the following: given two families of regions R and B in the plane, each of which consists of connected, nonpiercing regions, the intersection hypergraph HR(B) = (B, {Br}r∈R), where Br = {b ∈ B : b ∩ r 6= ∅} has a planar support. Further, such a planar support can be computed in time polynomial in R, B, and the number of vertices in the arrangement of the regions in R ∪ B. Special cases of this result include the setting where either the family R, or the family B is a set of points. Our result unifies and generalizes several previous results on planar supports, PTASs for packing and covering problems on nonpiercing regions in the plane and coloring of intersection hypergraph of nonpiercing regions. 2012 ACM Subject Classification Mathematics of computing → Approximation algorithms Acknowledgements Part of the work was done when the first author was visiting NYU Abu Dhabi, and the author thanks the institution for its hospitality.
and phrases Geometric optimization; packing and covering; nonpiercing regions
Introduction
Hypergraphs arise naturally in several applications and it is often helpful to capture the
structure of the hypergraph using a sparse graph. One popular way to capture the structure
of a hypergraph is to construct a planar support, which is a planar graph G with the same
vertex set as the hypergraph such that every hyperedge induces a connected subgraph of G.
In this paper, we study hypergraphs that arise in several, primarily geometric settings. For
example, a set of points P , and family of disks D in the plane define a hypergraph, H(P, D),
where each disk d ∈ D defines a hyperedge P ∩ d. This is a widely studied hypergraph, and it
is well known that the Delaunay graph of the points is a planar support for this hypergraph.
Another hypergraph on the same objects is H(D, P ), where the vertices are the disks, and
each point p ∈ P defines a hyperdge {d ∈ D : d 3 p}. We refer to H(P, D) as the primal
hypergraph, and H(D, P ), as the dual hypergraph. A hypergraph that generalizes both these
hypergraphs is the following: Given a family R of red disks, and a family B of blue disks,
we define the intersection hypergraph H(B, R) as follows: the disks B are the vertices, and
each disk r ∈ R defines a hyperedge Br = {b ∈ B : r ∩ b =6 ∅}. Both the primal and dual
hypergraphs can be seen as special cases of this intersection hypergraph, by viewing the
points in the arrangement as either blue disks, or red disks of radius zero. Our result implies
that this intersection hypergraph has a planar support.
If instead of disks, we use topological generalizations like pseudodisks or kadmissible
regions, the primal hypergraph defined above still admits a planar support [13]. However,
it was not known if the same is true for the dual hypergraph. Our result implies that the
intersection hypergraphs of kadmissible regions (which includes pseudodisks) admits a planar
support which in particular means that the dual hypergraph has a planar support.
We go beyond kadmissible regions, and show that these results are true even if the regions
have holes (i.e., they need not be simply connected1). Previous proofs relied heavily on the
assumption that the regions are simply connected. In addition, our results are constructive
and give polynomial time algorithms. Thus, even for the primal hypergraph of kadmissible
regions, where a planar support was known to exist, we make progress by giving a polynomial
time construction.
The existence of a planar support for the primal hypergraph for kadmissible regions
has a surprising algorithmic consequence: it implies that local search yields a PTAS for the
hitting set problem (the set cover problem for the dual hypergraph) [12] for kadmissible
regions. Since a planar support for the dual hypergraph for kadmissible regions was not
known, the authors in [1] had to construct another suitable graph in order to prove that
local search yields a PTAS for the set cover problem for the primal hypergraph. Similarly,
for the Dominating Set problem for kadmissible regions, an alternate graph construction
was required in [1], which was a generalization of a previous construction of [7] for disks.
We unify these results by considering a problem that generalizes all three problems: hitting
set, set cover, and dominating set for which our result implies a PTAS. Our result is in fact
stronger and is not implied by the previous results.
Prior to our result, Chan and HarPeled [4] proved that an arrangement of kadmissible
regions of depth two 2 admits a planar support. In fact, in this case, the intersection graph3
of the regions is itself the planar support. They used this result to obtain a PTAS for the
independent set problem for kadmissible regions [4]. Our result also implies PTASs for
generalized versions of packing problems considered in [1] and [6]. For some of the problems,
we obtain a PTAS where only a constant factor approximation was known.
In a different line of work, Keszegh [10], recently proved the following interesting result:
the intersection hypergraph of two families of pseudodisks is four colorable. This result
follows immediately from our result since planar graphs are four colorable. Keszegh’s paper
has several other results but the above result is the central tool to which most of the paper
is devoted and from which the other results follow.
The result of Keszegh extends the result of Keller and Smorodinsky [9] where the
hypergraph is defined by a single family of pseudodisks, the vertex set is the set of pseudodisks
and the hyperedges consists of open or closed neighborhood of each pseudodisk in the
intersection graph.
Before describing our results and other related results in more detail, we introduce the
necessary definitions and notation.
1 A region is simply connected if any loop can be continuously shrunk to a point while staying within the
region, which is true for a disk, but not for an annulus.
2 i.e., no more than two intersect at any point in the plane
3 An intersection graph on a set of regions is a graph whose vertices are the regions, and two vertices are
adjacent if their corresponding regions intersect.
Definitions and Notation
We will use the term region to refer to a set γ in the plane that can be described as
γ = γ¯ \ int(H1 ∪ · · · ∪ Hk), where H1, · · · , Hk are disjoint, compact, simply connected regions
contained in the compact, simply connected region γ¯. Essentially, γ is a compact connected
region with holes. We will refer to the region γ¯ as the filled region corresponding to γ, and
the regions H1, · · · , Hk as the holes in γ. We will refer to the boundary of γ¯ as the outer
boundary of γ.
We say that a set of regions are in general position if a) the boundaries of any two of the
regions intersect a finite number of times, and cross at these points, b) the boundaries of
three (or more) of the regions do not intersect at a common point, and c) any region can
be expanded by a small nonzero amount without changing the arrangement of the regions
combinatorially (see Definition 8 in Section 2 for a formal definition of expansion of a region).
Furthermore, we say that a set of points and a set of regions are together in general position
if none of the points lie on the boundary of any of the regions. In the rest of the paper,
we will always assume general position. Two regions γ1 and γ2 in the plane are said to be
nonpiercing if the sets γ1 \ γ2, and γ2 \ γ1 are both connected. A family Γ of regions is said
to be nonpiercing if the regions in Γ are pairwise nonpiercing.
I Remark. The term nonpiercing has been used previously to refer to a family of regions
defined exactly as we have except that the regions are required to be simply connected (i.e.,
not containing holes). The term kadmissible refers to such nonpiercing families where
in addition the boundaries of each pair of regions intersect at most k times. The term
pseudodisks is used to refer to a 2admissible family of regions. To be more consistent with
the literature, we should be using a term like “nonpiercing regions with holes”. However, for
better readability we stick to using a shorter term.
Given a set Γ of nonpiercing regions, and a set P of points in the plane, we define the
primal hypergraph H(P, Γ) as the hypergraph in which the vertex set is P , and there is a
hyperedge γ ∩ P corresponding to each γ ∈ Γ. We define the dual hypergraph H(Γ, P ) as
the hypergraph in which the vertex set is Γ, and corresponding to each p ∈ P , there is a
hyperedge {γ ∈ Γ : γ 3 p}. Finally, given two families of nonpiercing regions R and B, we
define their intersection hypergraph H(B, R) as the hypergraph in which the vertex set is B,
and corresponding to each region r ∈ R there is a hyperedge Br = {b ∈ B : r ∩ b 6= ∅}.
1.1
Our Results and Implications
The main result we prove in this paper is the following:
I Theorem 1. Given two families R and B of nonpiercing regions, the intersection
hypergraph H(B, R) admits a planar support.
We now describe the implications of Theorem 1. Due to shortage of space in this extended
abstract, we do not prove the claimed consequences of Theorem 1, as they follow in a
straightforward manner from standard arguments.
Generalized Set Cover Problem for nonpiercing regions
Given a family R of red nonpiercing regions, and another family B of blue nonpiercing
regions, such that each r ∈ R is intersected by at least one b ∈ B, find the smallest subset
B0 ⊆ B such that each r ∈ R is intersected by at least one b ∈ B0.
The following theorem below follows in a straightforward manner from the framework
in [12] using Theorem 1.
I Theorem 2. There is a PTAS for the Generalized Set Cover problem for nonpiercing
regions.
When B is a set of points in the plane, this problem is equivalent to the hitting set
problem for nonpiercing regions: given a set of points and a family of nonpiercing regions,
find the smallest subset of points such that each region contains at least one point from our
chosen subset of points. When R is a set of points in the plane, this problem is equivalent
to the set cover problem for nonpiercing regions in the plane: given a set of points and
nonpiercing regions in the plane, find the smallest subset of the regions so that each input
point is covered by one of the chosen subset of regions. When both B and R are identical
families of nonpiercing regions, the problem is equivalent to the dominating set problem
for nonpiercing regions: given a family of nonpiercing regions, find the smallest subset of
regions so that each of the other regions intersects at least one of the chosen regions. Thus a
PTAS for the generalized set cover problem for nonpiercing regions implies a PTAS for all
three problems: hitting set, set cover and dominating set. However, the reverse is not true:
the PTASs for these three problems together do not imply a PTAS for the generalized set
cover problem.
A PTAS for the hitting set problem for simply connected nonpiercing regions in the
plane, and halfspaces in R3 was given in [12]. For the set cover problem for disks in the plane
a PTAS follows from the result for halfspaces, via a standard lifting to three dimensions. For
the dominating set problem for disks in the plane, a PTAS was given in [7]. Generalizations
of the PTASs for the set cover and dominating set problems for disks to simply connected
nonpiercing regions in the plane were given in [1]. None of the earlier results work in the
setting where the regions are allowed to have holes.
Weighted Covering problems
In the weighted variant of the generalized set cover problem, each region b ∈ B has a
nonnegative weight, and the goal is to minimize the total weight of the chosen set B0 ⊆ B. Chan
et. al. [3], building on the work of Varadarajan [14], obtained constantfactor approximation
algorithms for set systems with linear shallowcell complexity 4. Note that the number of
hyperedges of size 2 in H(B, R) is O(B) since each such hyperedge corresponds to an edge
in the planar support which cannot have more than 3B − 6 edges. Since this is true for
the projection of the hypergraph on any subset of B, a standard probabilistic argument due
to Clarkson and Shor [5] implies that the number of hyperedges in H(B, R) of size at most
k is O(kn) implying that the shallow cell complexity of H(B, R) is linear. Our result thus
implies a constant factor approximation for the weighted variant of the generalized set cover
problem via the framework of Chan et. al. [3].
I Theorem 3. There is an O(1)approximation algorithm for the weighted Generalized Set
Cover problem for nonpiercing regions.
Generalized Set Packing problem for nonpiercing regions
Given a set R of red nonpiercing regions, and a set B of blue nonpiercing regions where each
red region has a capacity bounded above by a constant C > 0, find a maximum cardinality
subset B0 ⊆ B, such that the number of blue regions in B0 intersecting any r ∈ R does not
exceed the capacity of the region r.
4 See [3] for the definition of Shallow Cell Complexity.
The theorem below follows in a straightforward manner from the framework in [1] using
Theorem 1.
I Theorem 4. There is a PTAS for the generalized set packing problem for nonpiercing
regions, when each region has a capacity bounded above by a constant.
The generalized set packing problem specializes to the region packing problem when the
set R is a set of points: Given a family B of nonpiercing regions, and a set R of points, each
with a capacity bounded above by a constant C find a maximum subset B0 ⊆ B, such that no
point r ∈ R is covered by more than its capacity. When C = 1, note that the region packing
problem is the discrete independent set problem which itself generalizes the independent set
problem in the intersection graph of the regions in B. In [4], Chan and HarPeled gave a
PTAS for the independent set problem for a set of simply connected, nonpiercing regions in
the plane. When the set B is a set of points, the generalized set packing problem specializes
to the point packing problem: Given a set of points B, and a family R of nonpiercing regions,
each with capacity upper bounded by a constant C, find a maximum cardinality subset of
points B0 ⊆ B, such that the number of points of B0 in any r ∈ R is at most the capacity r.
In [1], a PTAS was given for the region packing problem, when the regions were assumed
to be simply connected nonpiercing regions where the boundaries of each pair of regions
intersect at most a constant number of times. For the point packing problem, again [1]
gave a PTAS when C = 1. For larger values of C, only constantfactor approximation
algorithms were known [6]. Our result thus extends the PTAS in [1] for any constant C.
Furthermore, our PTAS for all the above problems works for nonpiercing regions with holes,
which generalizes the earlier results, but does not follow from them.
I Remark. All the PTASs mentioned above follow a local search framework, which requires
the construction of a suitable graph for analysis. In earlier work, the graph construction
for all the problems above relied on the fact that the regions were simply connected, and
nonpiercing, and did not extend to regions with holes. As far as we are aware, the above
problems were not studied for nonpiercing regions with holes. Further, each problem required
a different graph construction. It is satisfying to finally have a unified view of all these
problems.
Hypergraph Coloring
Recently, Keszegh [10], proved the following result which generalizes the result of Keller
and Smorodinsky [9]: Let R and B be two families of pseudodisks. Then, the intersection
hypergraph H(B, R) admits a coloring with 4 colors, and a conflictfree coloring with O(log n)
colors. Keszegh’s result follows from our result due to the fact that the planar support of
H(B, R) is four colorable, and a valid coloring of the planar support is a valid coloring of the
hypergraph. Our result thus extends Keszegh’s result to nonpiercing regions. In order to
prove his result, Keszegh proves that the Delaunay graph, G = (B, E) where the vertex set
is B and E is the set of hyperedges of size 2 in H(B, R), is planar. Observe that our result
is stronger since every edge in the Delaunay graph must be in the planar support.
I Theorem 5. Given two families R and B of nonpiercing regions, the intersection
hypergraph H(B, R) can be colored with at most 4 colors.
β
p2
α
α
β
In this section, we define cell bypassing5, a basic operation that is used to simplify a given
arrangement of nonpiercing regions. Let Γ denote a set of nonpiercing regions, and let
A denote the arrangement of these regions. We will show that whenever there is a point
contained in at least 3 regions, we can simplify the arrangement by modifying a region γ ∈ Γ,
while maintaining key properties required to construct a planar support.
For a region γ, we let ∂γ denote the boundary of γ, and int(γ) to denote the interior of
γ. For a family of regions Γ, define ∂Γ = Sγ∈Γ ∂γ where ∂γ denotes the boundary of the
region γ. The closure of each connected component of R2 \ ∂Γ defines a cell. For a cell C, we
define the following: ΓC denotes the set of regions containing the cell C, depth(C) denotes
ΓC , and ∂C denotes the boundary of C. If an arc of ∂γ lies on ∂C, then the region γ is
said to contribute to ∂C. We define degree(C) as the number of arcs on the boundary of C.
Note that the same region may contribute multiple arcs to ∂C.
We say that two cells are adjacent if their boundaries share an arc of positive length. We
define the cell adjacency graph 6 GΓ of Γ as the graph in which vertices correspond to the cells
and two vertices are adjacent in the graph if the corresponding cells are adjacent. Clearly,
GΓ is a planar graph. Observe that the degree of a cell in GΓ is equal to the number of arcs
on its boundary. Also note that the depth of adjacent cells C and C0 in an arrangement of
regions differ by exactly 1. If depth(C) < depth(C0), then ΓC0 = ΓC ∪ {ρ}, where ρ is the
unique region such that the arc C ∩ C0 ⊆ ∂ρ. We say that a cell C is maximal if its depth is
more than the depth of any cell C0 adjacent to it.
I Lemma 6. All regions in Γ contributing to the boundary of a maximal cell C in A contain
the cell C.
Proof. For contradiction, assume that there is an arc α of ∂γ that appears on ∂C, but
C 6⊆ γ. Let C0 be the cell adjacent to C along α. Then, depth(C0) > depth(C) as all regions
containing C also contain C0, and C0 is additionally contained in γ. This contradicts the
maximality of C. J
5 The notion of cell bypassing is similar is spirit to the notion of lens bypassing in [1] and to the notion of
core decomposition in [11]. However, the technical difference is critical for the applications in this paper.
6 The cell adjacency graph is the geometric dual of the arrangement graph in which the intersection
points of the boundaries of the regions are the vertices and two vertices are adjacent in the graph if
they appear consecutively along the boundary of some region.
I Lemma 7. Let C be a maximal simply connected cell in A, whose boundary arcs are
labelled by the regions in Γ contributing them. Let σ be the cyclic sequence of the labels of
the arcs in counterclockwise order along ∂C. Then, σ is a Davenport Schinzel sequence of
order 2 i.e., it does not contain a subsequence of the form α, β, α, β.
Proof. For contradiction, let a1, b1, a2, b2 be four arcs appearing in cyclic order along ∂C,
where a1 and a2 have the label α, and b1 and b2 have the label β. Since C is a maximal
cell, note that C ⊆ α ∩ β. Let p1 and p2 be points in the interior of the arcs a1 and a2
respectively. Similarly, let q1 and q2 be points in the interior of the arcs b1 and b2 respectively.
See Figure 1. Since q1 and q2 lie on the boundary of α \ β which by assumption is connected,
there is a curve τ joining q1 and q2 whose interior lies in α \ β. Similarly, there is a curve
τ 0 joining p1 and p2 whose interior lies in β \ α. Note that the interiors of neither τ , nor
τ 0 intersect C, since C does not intersect either α \ β or β \ α. Since p1, q1, p2, q2 appear
along ∂C in that order, and since C does not have any holes (i.e., it is simply connected), τ
and τ 0 must intersect at a point outside C and in the interior of both the curves. This is a
contradiction since α \ β and β \ α are disjoint sets. J
I Definition 8 ( expansion). For any region R, define an expansion of R, denoted R as
the Minkowski sum of R and a ball of an arbitrarily small radius centered at the origin.
I Remark. The expansion of a region is necessary for technical reasons. The parameter
is always chosen to be small enough so that combinatorial structure of the arrangement
does not change if a region R is replaced by R in the family Γ. Due to general position
assumptions, such an always exists. The choice of thus depends on the other regions in
the family Γ, but we supress this dependency for better readability.
I Definition 9 (Good region). For a maximal cell C in A, a region γ ∈ Γ is said to be a good
region for C if the following conditions hold: i) γ contributes to the boundary of C and ii)
(Γ \ {γ}) ∪ {γ0}, where γ0 = γ \ int(C ), is a nonpiercing family.
I Lemma 10. For any maximal simply connected cell C in A, there is a region γ ∈ Γ that
contributes exactly one arc to the boundary of C. Furthermore, any such region is good for
C.
Proof. First we argue that there is a region that contributes exactly one arc to ∂C. If every
arc on ∂C has a distinct label, we are done. Otherwise, consider two arcs a1 and a2 having
the same label α that are closest to each other in counterclockwise order along ∂C. Let b
be any arc lying between a1 and a2. There is such an arc since consecutive arcs along ∂C
cannot have the same label. Let γ be the label of the arc b. By the choice of a1 and a2 and
Lemma 7 there cannot be any other arc on ∂C with label γ. Thus, there is at least one
region γ that contributes exactly one arc to ∂C.
We now show that any such region γ is a good region for C. The fact that γ contributes
exactly one arc to ∂C, along with the fact that C is simply connected implies that γ0 =
γ \ int(C ) is connected. For any other region ν ∈ Γ, we now argue that both ν \ γ0 and
γ0 \ ν are connected. Suppose first that ν does not contain C. Then ν \ γ0 = ν \ γ which is
connected. Also γ0 \ ν = (γ \ ν) \ int(C ), which is connected since the boundaries of γ \ ν
and C intersect only on one arc along ∂C. Now suppose that ν contains C. Then note that
γ0 \ ν is almost the same as γ \ ν. In fact if ν does not contribute to the boundary of C, then
γ0 \ ν = γ \ ν. Otherwise, γ0 \ ν is obtained by shaving off a thin strips of width from the
boundary of γ \ ν. Thus γ0 \ ν is connected. Also ν \ γ0 = (ν \ γ) ∪ C . Since ν \ γ and C
are both connected, and have a nonempty intersection, their union is also connected. J
I Lemma 11. Any maximal cell C in A of depth at least two, and containing a hole, has
exactly one hole H. Exactly one region γ ∈ Γ contributes to the boundary of H 7. This region
γ does not contribute any other arc to the boundary of C, and is a good region for C.
Proof. First, observe that the boundary of any hole in C can be contributed to by at most
one region. To see this assume to the contrary that two or more regions contribute to the
boundary of a hole in C. If the boundaries of two of these regions intersect on the boundary
of the hole, then the boundaries of those regions intersect the interior of the cell, which is
impossible. Otherwise the hole belongs to at least two distinct regions, which violates the
general position assumption.
Let Hγ be a hole in C whose boundary is contributed to by the region γ. We will show
that Hγ is the only hole in C. If γ contributed to another hole, say Hγ0 in C, then for any
other region β (such a region exists, since depth(C) ≥ 2) containing C intersects int(Hγ), as
well as int(Hγ0), and thus β \ γ cannot be connected. Thus γ cannot contribute to any other
hole in C.
Let ρ be any other region containing C. Then, note that ρ intersects int(Hγ) and since
C is contained in both γ and ρ, we must have ρ \ γ ⊂ int(Hγ), as otherwise ρ \ γ would not
be connected. This implies that ρ ⊂ int(γ ∪ Hγ). In particular, this means that ρ ⊂ int(γ¯),
where γ¯ is the outer boundary of the region γ. If ρ contributed to the boundary of another
hole Hρ in C, then by the same argument, we would have γ ⊂ int(ρ¯), a contradiction. This
means that Hγ is the only hole in C.
Since the depth of C is at least two, there is at least one other region ρ containing C.
However, as argued before ρ ⊂ int(γ ∪ Hγ). Since C ⊆ ρ the boundary of γ cannot contribute
any other arc (other than the boundary of Hγ) to the boundary of C.
We now argue that γ is a good region for C. Let γ0 = γ \ C . The region γ0 is connected
since γ0 is obtained from γ by replacing the hole Hγ by a larger hole H = C ∪ Hγ ⊂ int(γ¯) 8
containing Hγ. H is simply connected since Hγ is the only hole in C. Also note that no
other hole of γ intersects H. Let ν be any other region in Γ. We will show that both ν \ γ0,
and γ0 \ ν are connected. Suppose first that ν ∩ C = ∅. Then, ν \ γ0 = ν \ γ, which by
assumption, is connected. Also, note that γ0 \ ν is obtained by replacing the hole Hγ in γ \ ν
by the larger hole H which contains Hγ, is contained in int(γ¯), and does not intersect any
other holes in γ \ ν. Thus γ0 \ ν is connected. Now suppose that C ⊂ ν. Then γ0 \ ν is almost
the same as γ \ ν and is obtained by shaving off a thin strip of width from the boundary of
γ \ ν. Thus γ0 \ ν is connected. Also, ν \ γ0 = (ν \ γ) ∪ C . Since ν \ γ intersects C near the
boundary of the hole Hγ, ν \ γ0 is connected. J
Lemmas 10 and 11 imply the following lemma.
I Lemma 12. Any maximal cell C of depth at least two in the arrangement A has a good
region.
I Definition 13 (Cell Bypassing). Let C be a maximal cell and γ be a good region for C.
Then, by cellbypassing of C by γ, we mean the modification of γ to γ0 = γ \ int(C ). See
Figures 2a and 2b.
The following observations are immediate.
7 In other words H is a hole of γ.
8 Recall that γ does not contribute to the outer boundary of C.
(a) Case 1: C is simply connected. The
portion of the boundary of γ0 distinct from
γ is shown in red.
(b) Case 2: C is not simply connected. The
hole Hγ of γ is expanded to H, so that it
now contains C.
I Observation 14. If a maximal cell C is identical to a region γ, then γ is good for C and
cellbypassing replaces γ by an empty region, effectively removing γ from the family of regions.
I Observation 15. When a cell C of depth d is bypassed by a region γ, the number of cells
of depth d decreases by 1. The number of cells of lower depth may increase. In fact, there
are at most O(Δ) newly created cells of depth either d − 2 or d − 3, where Δ is the degree of
the cell C.
3
Construction of a planar support
We first give the construction of a planar support for the dual hypergraph defined by a set
of nonpiercing regions, and all points in the plane. We then use this to construct a planar
support for the intersection hypergraph of two nonpiercing families.
3.1
Planar Support for Dual Hypergraph
In the following, let Γ be a family of nonpiercing regions and let A denote their arrangement.
We construct a planar support for the hypergraph H(Γ, R2) 9, which immediately implies
a planar support for any P ⊆ R2. For a set of S ⊆ Γ of the regions and any point p ∈ R2,
we denote the set of regions in S that contain p by Sp. Similarly for a cell C, SC denotes
the set of regions in S containing C. For a graph G and a subset U of the vertices of G, we
denote the subgraph of G induced by U by G[U ].
I Lemma 16. Let C be a maximal cell in A with depth(C) ≥ 3. Let Γ0 be the arrangement
obtained by cell bypassing of C by a good region γ of C. Then, a planar support for H(Γ, R2)
can be constructed from a planar support for H(Γ0, R2).
Proof. Suppose ∂C consists of a single arc. Then, C is identical to some region γ ∈ Γ. Since
depth(C) ≥ 3, γ is completely contained in another region ρ ∈ Γ. Bypassing C in this case
results in the arrangement Γ0 = Γ \ {γ}. Given a planar support G0 for H(Γ0, R2), we can
obtain a planar support G for H(Γ, R2) by simply adding a new vertex for γ and an edge
between γ and ρ. Since we obtain G by adding a vertex of degree 1 to the planar graph
G0, G is planar. To see that G is a support, consider any point p in the plane. If p ∈/ γ
then Γp = Γ0p and therefore G[Γp] = G0[Γ0p] which is connected. On the other hand, if p ∈ γ,
9 While there are infinitely many points, the hypergraph is still finite, since all points in the same cell of
the arrangement A define the same hyperedge.
then Γp = Γ0p ∪ {γ}. Since G0 is a planar support for Γ0, G[Γ0p] = G0[Γ0p] is connected. Since
ρ ∈ Γ0(p) and there is an edge between γ and ρ in G, it follows that G[Γ(p)] is connected.
Now, suppose ∂C contains at least two arcs. Let ρ 6= γ be a region that contributes to
∂C. Then, in the arrangement of Γ, there is a cell X adjacent to C such that ΓX = ΓC \ {ρ}.
After γ is modified to bypass C, in the arrangement of Γ0, there is still a cell X0 such that
Γ0X0 = ΓX and there is a cell C0 s.t. Γ0C0 = ΓC \ {γ}. Since ΓC  ≥ 3, the sets Γ0X0 and Γ0C0
intersect. Also note that their union is ΓC . Since G0[Γ0X0 ] and G0[Γ0C0 ] are connected, it
follows that G0[ΓC ] is connected. Thus G = G0 is a planar support for H(Γ, R2). J
The following lemma follows from [4]. The intersection graph of a set of regions is a graph
in which the vertices correspond to the regions and edges correspond to pairs of intersecting
regions.
I Lemma 17 ([4]). If Γ is a family of nonpiercing regions so that no cell in their arrangement
has depth more than 2, then the intersection graph of Γ is planar, and is a support for
H(Γ, R2).
I Remark. Even though the proof presented in [4] is for kadmissible regions, essentially the
same proof works for nonpiercing regions (with holes).
I Theorem 18. Let Γ be a family of nonpiercing regions. Then, there exists a planar
support for H(Γ, R2).
Proof. For any set of regions Γ, define dΓ to be the maximum depth of any cell in the
arrangement of Γ, and nΓ to be the number of cells in the arrangment with depth dΓ. We
use induction on the pair (dΓ, nΓ). If dΓ ≤ 2, then by Lemma 17, the intersection graph of
Γ is the required planar support. Given a set of regions with dΓ ≥ 3, let Γ0 be the set of
regions obtained after bypassing a cell of maximum depth in the arrangement of Γ. Then,
by Observation 15, the pair (dΓ0 , nΓ0 ) is lexicographically smaller than (dΓ, nΓ). Inductively,
a planar support for H(Γ0, R2) exists, which by Lemma 16, is also a planar support for
H(Γ, R2). J
3.2
Planar Support for the Intersection Hypergraph
Given a family R of red nonpiercing regions, and a family B of blue nonpiercing regions, we
prove that there exists a planar support for the intersection hypergraph H(B, R) = (B, E),
where E = {Br : r ∈ R}, and Br = {b ∈ B : b ∩ r 6= ∅}. However, it is not essential for the
rest of the paper. Note that we can assume without loss of generality that any red region
intersects at least one blue region, and we make this assumption throughout this section.
This implies that any maximal cell in the arrangement of R ∪ B has depth at least 2.
Let B r = {b ∩ r : b ∈ Br} be the set of regions obtained by intersecting the regions in B

with the region r. A region in B r may have multiple connected components, but we still

treat it as a single region. Let G r(B) be the intersection graph of these regions. We start

with the following simple observation.
I Lemma 19. If for each red region r ∈ R, the graph G r(B) is connected, then the planar

support for the hypergraph H(B, R2), is also a planar support for the hypergraph H(B, R).
Proof. Let G be the planar support of the hypergraph H(B, R2) defined by the blue regions.
Recall that the graph G guarantees that for each p ∈ R2, the set of blue regions containing p
induce a connected subgraph of G. Consider any red region r ∈ R. Since, by assumption,
the graph G r(B) is connected, for any pair of blue regions s, t ∈ Br intersecting r, there is

a sequence s = b1, · · · , bk = t, of regions in Br such that adjacent regions bi, bi+1 intersect
at a point qi ∈ r. This means that there is a path in G between bi and bi+1 via the regions
containing qi, i.e., via regions in Br. This in turn implies that there is a path between s and
t in G via regions in Br. J
The construction of the planar support in the general setting is a reduction to the setting
in Lemma 19.
I Lemma 20. Given a set of red and blue nonpiercing regions R and B, we can obtain a
modified set of red regions R0, such that (i) the set of regions R0 is nonpiercing, (ii) in the
arrangement of R0 ∪ B, each maximal cell is contained in some blue region in B, and (iii)
the intersection hypergraph H(B, R0) is isomorphic to H(B, R).
Proof. Consider a maximal cell C in the arrangement of R ∪ B that is not contained in
any blue region. Then, C is also a maximal cell in the arrangement of R. Since R is a
nonpiercing family, by Lemma 12 we can bypass C. This does not change the intersection
hypergraph of the red and blue regions, since no redblue intersection is lost or gained as a
consequence of bypassing the cell C. We repeat this process until each maximal cell is in
some blue region. The modified set of regions thus obtained satisfy the conditions of the
lemma. J
I Lemma 21. Given two families of red and blue nonpiercing regions R and B, such that
each maximal cell in the arrangement A of R ∪ B is contained in some region b ∈ B, we
can add a fake blue region βC corresponding to each cell C in A that is not contained in
any blue region in B, such that R, B0 and B00, where B00 is the set of fake blue regions and
B0 = B ∪ B00, satisfy: (i) B0 is a family of nonpiercing regions, (ii) each βC intersects only
those regions in R that contain C, and (iii) for each r ∈ R, the intersection graph G r(B0) is

connected.
Proof. Let C be a cell not contained in any of the blue regions. We define the fake blue
region for this cell as βC = C \ Sr∈R, C6⊂r int(rδ), where δ < and is sufficiently small.
See Figure 3. Intuitively, βC is roughly the same as C but we modify its boundary slightly
so that it intersects all the blue regions that contribute to the boundary C but does not
intersect any red regions not containing C. Defining βC this way ensures that property (ii)
in the statement of the Lemma is satisfied. Choosing to be sufficiently small also ensures
that property (i) is satisfied. Finally, choosing δ < also ensures that each red region in R
is covered by the union of the blue regions in B0 which implies property (iii). To see this,
consider a point p contained in r ∈ R, and let D be the cell in A containing p. If p is not
contained in a blue region in B, then since D does not lie in any blue region, we add a fake
blue region βD corresponding to D. If p lies at a distance of at least δ from any arc of ∂D,
contributed by a red region in R not containing D, then note that p lies in βD. Otherwise, if
p lies within distance δ of some arc α contributed by a red region not containing D, the cell
D0 adjacent to D sharing the arc α is also not contained in any blue region in B. In this
case, since > δ, βD0 contains p. This implies that Gr(B) is connected for each r ∈ R. J
We are now ready to prove Theorem 1.
Proof of Theorem 1. If G r(B) is connected for each r ∈ R, then we obtain a planar support

by Lemma 19. If not, let R0 be the set of modified red regions obtained by applying Lemma
20. Let B00 be the set of fake blue regions added by applying Lemma 21 to the blue regions
B, and the modified red regions R0. Now, by Lemma 19, a planar support for H(B0, R2) is a
planar support for H(B0, R0). Let G0 be this planar support. We show that we can obtain a
planar support G for H(B, R0), by suitably modifying G0. G is also a support for H(B, R)
since H(B, R) is isomorphic to H(B, R0). In the following, we refer to a vertex in the support
graph by the corresponding region. Let A be the arrangement of B ∪ R0, and let B0 denote
the arrangement of the regions in B0. Let C be a cell in A not contained in any blue region
in B. By Lemma 20, C is not a maximal cell. Thus, there is a fake blue region βC ∈ B0
corresponding to C. Since C is not maximal, we can pick a cell C0 (arbitrarily chosen in
case of ties) adjacent to C in A, such that depth(C0) = depth(C) + 1. Let b be a fake or real
blue region defined as follows: if C0 is not contained in any blue region in B then b = βC0 ,
otherwise b is the unique blue region in B containing C0.
Note that b forms a depth 2 intersection with βC in B0 (i.e. there is a point in the plane
contained in just these two regions), and therefore, βC and b are adjacent in G0. We orient
the corresponding edge in G0 from βC to b. Thus, for each fake region in G0, there is exactly
one outgoing oriented edge incident on it. Note that not all edges in G0 are oriented.
By Property (ii) of Lemma 21, any red region intersecting βC contains C. Such a region
also contains C0 since C and C0 are adjacent cells in A, and depth(C0) = depth(C) + 1. Thus
all red regions intersecting βC also intersect b. In other words, the set of red regions in R0
intersecting βC is a subset of the red regions in R0 intersecting b. This is a key property we
will use later. By Lemma 20, each maximal cell in A is contained in a blue region in B, and
therefore there is a unique directed path starting from a fake region βC and ending at a real
blue region ˜b ∈ B. Crucially, the set of red regions intersecting any fake region βC is a subset
of the red regions intersecting ˜b due to the key property mentioned earlier.
The set of oriented edges in G0 form a spanning forest, where each arc is oriented towards
the root of a tree, and the root of each tree corresponds to a real region. We obtain G
by contracting the edges in the forest (i.e., all oriented edges in G), effectively merging all
nodes in a subtree to its root. Since edge contraction preserves planarity, it follows that
G is planar. To see that G is a support for H(B, R0), let b, b0 ∈ B be a pair of blue regions
intersecting a red region r ∈ R0. Since G0 is a planar support for H(B0, R0), there is a path
b = b1, . . . , bk = b0 such that each region bi in the path intersects r. Since each fake region
on the path is merged with a real region that also intersects r, it follows that there is a path
in G between b and b0, such that all regions along the path intersect r.
Finally, by Lemma 20, since H(B, R0) is isomorphic to H(B, R), G is the desired planar
support. J
3.3
Algorithms
The proofs for the existence of the planar supports given in Section 3 are constructive and
can be converted to an algorithm with running time O(n3 + m) where n is the number of
regions and m is the number of vertices in the arrangement of the regions. Note that if the
boundaries of every pair of regions intersect at most O(n) times, then m is O(n3). While
the applications mentioned in this paper require only the existence of a planar support, the
algorithmic question is natural and may have applications in hypergraph visualization (see [2]
and [8]). In the proofs of the existence of the planar support, we modify the arrangement of
the regions by doing cell bypassing. For algorithmic purposes, instead of maintaining the
geometric shapes of the regions, we can maintain the dual arrangement graph. In addition,
we orient the edges from a cell of lower depth to a cell of higher depth and label the edge by
the index of the region whose boundary separates the cells. The vertex corresponding to a
cell also stores the depth of the cell and pointers to neighboring vertices. The size of the
graph is O(m). We assume that the initial dual arrangement graph H of the input set of
region Γ is given. It is not at all apparent that these algorithms run in polynomial time, since
the number of cells in the arrangement can increase after each cellbypassing. However, using
a more detailed combinatorial analysis, we can define a suitable potential function on the
arrangement of the regions, so that each time we do a cell bypassing operation the potential
goes down by an amount proportional to the time spent in the cellbypassing operation.
The initial potential can be shown to be O(n3 + m) which implies the upper bound on the
running time. The details of the analysis are not included due to lack of space.
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